Basic concepts of organic chemistry

In a nutshell

Organic chemistry is the chemistry of carbon compounds. This subtopic gives you the toolkit every later organic topic uses: how to name a compound, how to draw it, how to sort compounds into homologous series, how to spot structural isomers, and how to show bonds breaking and forming with curly arrows.

Get this right and the alkanes, alkenes, alcohols and haloalkanes notes become far easier.

Assumed knowledge: Amount of substance (empirical and molecular formulae), Bonding and structure (covalent bonds, lone pairs, electronegativity and polar bonds).

Core content

Naming organic compounds: IUPAC rules

IUPAC names are systematic: each name describes exactly one structure, so there is no ambiguity.

A name is built from three parts:

• a stem for the number of carbons in the longest carbon chain
• a suffix for the main functional group (with its position number)
• prefixes for branches (alkyl groups) and halogens (with their position numbers)

Stems: the first ten alkanes and their alkyl groups (you must know all ten)

A ring of carbons takes the prefix cyclo-, e.g. cyclohexane.

Functional groups you need to name at AS

The naming method, step by step

1. Find the longest continuous carbon chain that contains the functional group. It does not have to be drawn in a straight line.
2. Number the chain from the end that gives the suffix group the lowest number. If there is no suffix group, give the branches and halogens the lowest numbers.
3. Name each branch or halogen with its position number.
4. Put the prefixes in alphabetical order (bromo before chloro before methyl). Ignore di- and tri- when sorting.
5. Use di-, tri-, tetra- for two, three or four identical groups, and give every one a number: 2,2-dichloropropane.

Punctuation and spelling rules

• Commas between numbers, hyphens between numbers and letters, no spaces: 2-methylbutan-1-ol.
• Drop the final e of the stem when the suffix starts with a vowel: propan-1-ol, propanal, propanone, propanoic acid.
• Keep the e when the next part starts with a consonant: ethane-1,2-diol.
• Aldehydes and carboxylic acids need no number: the group is always on carbon 1.
• An alkene's number is the lower of the two carbons in the C=C: CH₃CH=CHCH₃ is but-2-ene.

So CH₃CH₂CH(CH₃)CH₂OH is 2-methylbutan-1-ol: a four-carbon chain, OH on carbon 1, methyl branch on carbon 2.

Naming at AS is limited to the functional groups in the table above.

Naming and representing: types of formula

You must be able to read and write every type of formula, and choose the right one for the question.

Structural formula conventions

• Write the carboxyl group as COOH.
• Write an aldehyde group as CHO, never COH (COH reads as an alcohol).
• Brackets show a branch on the carbon before them: CH₃CH(CH₃)CH₃.
• R can stand for an alkyl group, or any part of a molecule not involved in the reaction.

Reading a skeletal formula

• Every line end and every corner is a carbon atom.
• Each carbon has enough hydrogen atoms (not drawn) to make four bonds.
• Atoms other than C and H on carbon are written in, including the H of an OH group.
• A double line is a C=C (or C=O) double bond.

OCR uses these symbols for cyclohexane and benzene:

Functional groups: homologous series and general formulae

A homologous series is a series of organic compounds having the same functional group but with each successive member differing by CH₂.

A functional group is a group of atoms responsible for the characteristic reactions of a compound. Because every member of a series has the same functional group, they all react in the same way.

An alkyl group has the formula C_nH_2n+1 (methyl CH₃, ethyl C₂H₅, and so on). It is an alkane with one H removed.

General formulae: predict any member of a series

To predict a member: substitute the number of carbons for n. Each step along the series adds one CH₂, so the M_r of an alkane rises by 14.0 each time.

Notice that alkenes and cycloalkanes share C_nH_2n. A general formula alone does not tell you the series.

Functional groups: aliphatic, alicyclic and aromatic

These three words describe the shape of the carbon framework.

• Aliphatic: carbon and hydrogen joined in straight chains, branched chains or non-aromatic rings. Examples: hexane, 2-methylpentane, cyclohexane.
• Alicyclic: an aliphatic compound arranged in non-aromatic rings, with or without side chains. Examples: cyclohexane, methylcyclohexane, cyclohexene.
• Aromatic: a compound containing a benzene ring.

So every alicyclic compound is also aliphatic. Cyclohexane is alicyclic (a ring) but not aromatic (no benzene ring).

Functional groups: saturated and unsaturated

• Saturated: single carbon-carbon bonds only.
• Unsaturated: multiple carbon-carbon bonds are present: C=C, C≡C or an aromatic ring.

The test is about carbon-carbon bonds only.

Isomerism: structural isomers

Structural isomers are compounds with the same molecular formula but different structural formulae.

The atoms are the same; how they are joined is different. Isomers can differ in:

• the carbon chain (straight or branched): pentane and 2-methylbutane
• the position of a functional group: propan-1-ol and propan-2-ol
• the functional group itself: propanal (aldehyde) and propanone (ketone), both C₃H₆O; hex-1-ene and cyclohexane, both C₆H₁₂

The OCR specification does not split these into named types: you only need the term structural isomers.

How to find every structural isomer of a molecular formula

1. Draw the longest straight chain first, and place any functional group in each different position.
2. Shorten the chain by one carbon, add a methyl branch, and repeat.
3. Keep shortening until no new structures appear.
4. Name every structure. Two structures with the same name are the same compound, so count it once.

Still don't get it? · is it really a different isomer?

Think of a bead necklace on a loose string. You can lay it straight, curl it into a U, or flip it over: it is still the same necklace. You only get a different necklace if you cut the string and thread the beads in a different order.

Molecules on paper are the same:

• A chain drawn bent, or drawn from right to left, is the same molecule. CH₃CH₂CH₂Cl and ClCH₂CH₂CH₃ are both 1-chloropropane.
• "3-chloropropane" is just 1-chloropropane numbered from the wrong end. Not new.
• A "branch" on the end carbon is not a branch at all: it just makes the chain longer. CH₃CH₂CH₂CH₂CH₃ drawn with the last carbon pointing down is still pentane, not "1-methylbutane".
• A genuinely new isomer means the atoms are connected in a different order: a new branch point, or the functional group on a different carbon.

The quickest check is to name each structure with the IUPAC rules. Different names mean different isomers. The same name means you have drawn one compound twice.

The exam version: structural isomers have the same molecular formula but different structural formulae.

Reaction mechanisms: homolytic and heterolytic fission

A covalent bond is a shared pair of electrons. When it breaks (bond fission), the two electrons can be shared out in two ways.

In heterolytic fission the more electronegative atom usually takes both electrons, so it becomes the negative ion. In H-Br, bromine takes the pair.

Reaction mechanisms: radicals

A radical is a species with an unpaired electron.

Radicals are shown with a dot, •, for the unpaired electron:

• chlorine radical: Cl•
• methyl radical: CH₃•

OCR rules for dots:

• A dot is required every time a species has a single unpaired electron, in every equation of a mechanism.
• A dot is not required for a diradical (two unpaired electrons), such as an oxygen atom, O.
• A radical has no charge. Cl• and Cl^- are different species.

Radical mechanisms are written as a sequence of equations, not with curly arrows. You will use this for the radical substitution of alkanes in Alkanes.

Reaction mechanisms: curly arrows and dipoles

A curly arrow shows the movement of an electron pair. It shows either heterolytic fission or the formation of a covalent bond.

The rules for drawing one

• Use a double-headed (full) arrowhead: it moves two electrons.
• The arrow starts from a bond, a lone pair of electrons or a negative charge. Never from an atom or a positive charge.
• The arrow ends where the pair goes:
  • pointing at the atom it will bond to, when a lone pair or negative charge forms a new covalent bond
  • on the atom that takes both electrons, when a bond breaks by heterolytic fission
• Include every relevant dipole: δ+ and δ- on polar bonds.

Half curly arrows (one electron) are not needed here.

The mechanism below is one you will learn properly in Haloalkanes. For now, look only at where each arrow starts and ends.

Reading Figure 6:

1. The C-Br bond is polar: Br is more electronegative, so the carbon is δ+ and the bromine δ-.
2. Arrow 1 starts at a lone pair on the O of OH^- and ends at the δ+ carbon: a new C-O bond forms.
3. Arrow 2 starts at the C-Br bond and ends on Br: the bond breaks by heterolytic fission, and Br leaves as Br^- with both electrons.

Still don't get it? · curly arrows

Think of a curly arrow as a delivery of a parcel that always holds two items. The tail of the arrow is where the parcel is picked up. The head is where it is dropped off. You can only pick a parcel up from somewhere that actually has one.

Now step by step:

• The "parcel" is an electron pair. Electron pairs exist in only three places: in a bond, as a lone pair on an atom, or as the extra electrons that give a negative charge. So the tail must start at one of those.
• A positive charge has lost electrons, so there is nothing to pick up there. An arrow never starts from a positive charge.
• If the pair is picked up from a bond and dropped on one of that bond's own atoms, that atom now owns both electrons. The bond has broken unevenly: that is heterolytic fission, and the atom becomes a negative ion.
• If the pair is picked up from a lone pair (or negative charge) and dropped on another atom, the two atoms now share it: a new covalent bond has formed.
• Electrons are attracted to the δ+ end of a polar bond, so arrows from a lone pair usually head for the δ+ atom. That is why the dipoles have to be drawn.

The exam version: a curly arrow shows the movement of an electron pair, showing either heterolytic fission or the formation of a covalent bond. It must start from a bond, a lone pair or a negative charge, and you must include the relevant dipoles.

Worked examples

Example 1: naming two compounds

Name CH₃CH₂CH(CH₃)CH(OH)CH₃.

1. Longest chain containing the OH carbon: 5 carbons, so the stem is pent-.
2. Number from the end that gives the OH the lowest number. From the right-hand CH₃: OH is on C2. From the left it would be C4, so number from the right.
3. With that numbering, the CH₃ branch is on C3.
4. The suffix starts with a vowel, so drop the e: pentan-2-ol.

Answer: 3-methylpentan-2-ol.

Name CH₃CHClCH(CH₃)CH₂CH₃.

1. Longest chain: 5 carbons, so pentane. There is no suffix group, so the chlorine and the methyl group are both prefixes.
2. Numbering from the left gives positions 2 (Cl) and 3 (CH₃). From the right gives 3 and 4. The first point of difference is 2 against 3, so number from the left.
3. Alphabetical order: chloro before methyl.

Answer: 2-chloro-3-methylpentane.

Where marks go: numbering from the wrong end (3-methyl-4-chloropentane loses the mark), listing the prefixes out of alphabetical order, or writing spaces ("2-chloro 3-methyl pentane").

Example 2: using a general formula

(a) Give the molecular formula of the alcohol with seven carbon atoms.

Alcohols are C_nH_2n+1OH. Substitute n = 7:

$\text{C}_7\text{H}_{(2 \times 7)+1}\text{OH} = \text{C}_7\text{H}_{15}\text{OH}$

Answer: C₇H₁₅OH (molecular formula C₇H₁₆O).

(b) An alkane has M_r = 72.0. Find its molecular formula. Use A_r: C = 12.0, H = 1.0 (given on the data sheet).

Write M_r of C_nH_2n+2 in terms of n:

$M_r = 12.0n + 1.0(2n + 2)$

$M_r = 14.0n + 2.0$

Set it equal to the given value and solve:

$14.0n + 2.0 = 72.0$

$n = \frac{72.0 - 2.0}{14.0} = 5$

Answer: C₅H₁₂ (check: 5 × 12.0 + 12 × 1.0 = 72.0).

Where marks go: forgetting the "+2" hydrogens and solving 14.0n = 72.0 (gives n = 5.14, which is not a whole number). If n does not come out as a whole number, your general formula is wrong.

Example 3: drawing all the structural isomers of C₄H₉Br

Follow the method: longest chain first, then branch.

1. Four-carbon chain, Br on the end carbon: CH₃CH₂CH₂CH₂Br.
2. Four-carbon chain, Br on an inner carbon: CH₃CH₂CHBrCH₃. (Br on C3 is the same as C2 numbered from the other end.)
3. Three-carbon chain with a methyl branch, Br on an end carbon: (CH₃)₂CHCH₂Br.
4. Three-carbon chain with a methyl branch, Br on the middle carbon: (CH₃)₃CBr.

Now name them to check there are no repeats:

Four different names, so C₄H₉Br has four structural isomers.

Where marks go: adding "3-bromobutane" or "1-bromo-1-methylpropane" as extra isomers. Both are just 2-bromobutane drawn differently.

Common exam mistakes

• Not using the longest chain. Calling CH₃CH₂CH(CH₂CH₃)CH₃ "2-ethylbutane"; the longest chain has five carbons, so it is 3-methylpentane. A "2-ethyl" or "1-methyl" in your answer means you have missed a longer chain.
• Numbering from the wrong end. The suffix group (OH, C=C) gets the lowest number first: CH₂=CHCH(CH₃)CH₃ is 3-methylbut-1-ene, not 2-methylbut-3-ene.
• Punctuation and the stem's e. "2-methyl butan-1-ol" (space), "2.3-dichloropentane" (full stop) and "propane-1-ol" (extra e) are all wrong. Write 2-methylbutan-1-ol, 2,3-dichloropentane, propan-1-ol.
• Prefixes not in alphabetical order. It is 2-bromo-3-chlorobutane, not 3-chloro-2-bromobutane.
• Writing an aldehyde as COH. CH₃COH looks like an alcohol; ethanal is CH₃CHO.
• Missing bonds in a displayed formula. "Displayed" means every bond, including the O-H bond. Writing -OH on the end, or joining the chain to the H of OH (C-H-O), loses the mark.
• Letters in a skeletal formula. Writing C or CH₃ at the corners of a skeletal formula, or forgetting that each line end is a carbon, so the carbon count is wrong.
• Same isomer twice. Drawing butane bent and calling it a new isomer, or listing 1-chloropropane and 3-chloropropane as different. Name each structure to check.
• Homologous series definition missing a part. It needs both "same functional group" and "each successive member differing by CH₂".
• Calling propanone unsaturated because of C=O. Saturated and unsaturated refer only to carbon-carbon bonds. Benzene, though, is unsaturated (aromatic ring).
• Homolytic and heterolytic swapped. Homolytic: each atom gets one electron, radicals form. Heterolytic: one atom gets both, ions form.
• Radical dot missing or wrong. Writing Cl instead of Cl•, or giving a radical a charge (Cl^-•). Every species with one unpaired electron needs its dot in every equation.
• Curly arrow from the wrong place. Starting an arrow at an atom (for example the O of OH^- rather than its lone pair) or at a positive charge, pointing it the wrong way, or leaving out the δ+/δ- dipole.

Key definitions

• Homologous series: a series of organic compounds having the same functional group but with each successive member differing by CH₂. (OCR requires this definition.)
• Functional group: a group of atoms responsible for the characteristic reactions of a compound.
• General formula: the simplest algebraic formula of a member of a homologous series, e.g. C_nH_2n+2 for an alkane.
• Structural formula: the minimal detail that shows the arrangement of atoms in a molecule.
• Displayed formula: shows the relative positioning of atoms and the bonds between them.
• Skeletal formula: the simplified organic formula, with hydrogen atoms removed from alkyl chains, leaving just a carbon skeleton and associated functional groups.
• Alkyl group: a group of formula C_nH_2n+1.
• Aliphatic: a compound containing carbon and hydrogen joined together in straight chains, branched chains or non-aromatic rings.
• Alicyclic: an aliphatic compound arranged in non-aromatic rings, with or without side chains.
• Aromatic: a compound containing a benzene ring.
• Saturated: single carbon-carbon bonds only.
• Unsaturated: the presence of multiple carbon-carbon bonds, including C=C, C≡C and aromatic rings.
• Structural isomers: compounds with the same molecular formula but different structural formulae.
• Homolytic fission: each bonding atom receives one electron from the bonded pair, forming two radicals.
• Heterolytic fission: one bonding atom receives both electrons from the bonded pair.
• Radical: a species with an unpaired electron.
• Curly arrow: the movement of an electron pair, showing either heterolytic fission or formation of a covalent bond.

Specification

• I can name organic compounds systematically using IUPAC rules, including branched chains and the AS functional groups. (4.1.1 a)
• I know the names of the first ten alkanes and their alkyl groups. (4.1.1 a)
• I can interpret and write general, structural, displayed and skeletal formulae, and choose the right one for the context. (4.1.1 b)
• I can recognise the OCR symbols for cyclohexane and benzene. (4.1.1 b)
• I can define a homologous series and use the terms functional group and alkyl group. (4.1.1 c i to iii)
• I can classify a compound as aliphatic, alicyclic or aromatic. (4.1.1 c iv to vi)
• I can classify a compound as saturated or unsaturated from its carbon-carbon bonds. (4.1.1 c vii)
• I can use a general formula to predict the formula of any member of a homologous series. (4.1.1 d)
• I can explain the term structural isomers and draw all the structural isomers of a given molecular formula. (4.1.1 e)
• I can describe homolytic and heterolytic fission in terms of where the bonded electrons go. (4.1.1 f)
• I can define a radical and use dots correctly for radicals in equations. (4.1.1 g)
• I can describe a curly arrow as the movement of an electron pair, showing heterolytic fission or bond formation. (4.1.1 h)
• I can draw curly arrows in a mechanism starting from a bond, a lone pair or a negative charge, with the relevant dipoles. (4.1.1 i)

Related notes

• Amount of substance
• Bonding and structure
• Alkanes
• Alkenes
• Alcohols
• Haloalkanes